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Maxima and Minima — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMaxima and Minima5 Marks
Question
A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle will produce the largest area of the window.
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Answer

Let the dimensions of the rectangular part be x and y.
Perimeter of the window = x + y + x + x + y = 12
⇒ 3x + 2y = 12
$\text{y}=\frac{12-3\text{x}}{2}....(\text{i})$
Area of the window $=\text{xy}+\frac{\sqrt{3}}{4}\text{x}^{2}$
$\Rightarrow\text{A}=\text{x}\Big(\frac{12-3\text{x}}{2}\Big)+\frac{\sqrt{3}}{4}\text{x}^{2}$
$\Rightarrow\text{A}=6\text{x}-\frac{3\text{x}^{2}}{2}+\frac{\sqrt{3}}{4}\text{x}^{2}$
$\Rightarrow\frac{\text{dA}}{\text{dx}}=6\text{x}-\frac{6\text{x}}{2}+\frac{2\sqrt{3}}{4}\text{x}$
For maximum or minimum values of A, We must have $\frac{\text{dA}}{\text{dx}}=0$
$\Rightarrow 6=\text{x}\Big(3-\frac{\sqrt{3}}{2}\Big) $
$\Rightarrow \text{x}=\frac{12}{6\sqrt{3}} $
Substituting the values of x in eq.(i), We get
$\Rightarrow \text{y}=\frac{12-3\Big(\frac{12}{6-\sqrt{3}}\Big)}{2} $
$\Rightarrow \text{y}=\frac{18-6\sqrt{3}}{6-\sqrt{3}}$.

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